Question

How do you find the equation of a circle passing through the points of intersection of the line x+y=4 and the circle x^2+y^2=16 and the point (10,2)?

Answer 1

`x^2+y^2-11x-11y+28=0`.

Explanation:

Method I :-

I will outline the soln.

First, solving the eqns. of the line and the circle, we will get the co-ords. of two pts. of intersection. Using these two pts. together with the given pt. `(10,2)`, we can find the eqn. of reqd. circle.

Method II :-

This Method is simpler than the Method I.

In this Method, we use the following Result :

Result :- If a circle `S : x^2+y^2+2gx+2fy+c=0` and a line

`L : ax+by+c=0` intersect, then the eqn. `S+kL=0, k in RR`

represents a circle passing through their pts. of intersection.

So, let us suppose that the reqd. eqn. of circle is,

`x^2+y^2-16+k(x+y-4)=0, k in RR`.

This circle passes thro. `(10,2)`. therefore,

`10^2+2^2-16+k(10+2-4)=0 rArr 88+k(8)=0.`

`rArr k=-11`. Hence, the reqd. eqn. is,

`x^2+y^2-16-11x-11y+44=0`, or,

`x^2+y^2-11x-11y+28=0`.

Enjoy Maths.!